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KCTF2021 第二题 南冥神功 wp
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发表于: 2021-5-12 10:08 5183
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迷宫 、 深度优先搜索
其中v10是求解关键,即走图path
根据深度优先搜索对每一个可能的分支路径深入到不能再深入为止,而且每个节点只能访问一次即获得路径
path坐标
path[]
根据path可以逆推出falg
sub_4AF840((
int
)&dword_4B8860,
"Input your code: "
);
sub_4B0AB0((
int
)&dword_4B8680, v25);
if
( strlen(v25) <
=
48
)
sub_4AF840((
int
)&dword_4B8860,
"Input your code: "
);
sub_4B0AB0((
int
)&dword_4B8680, v25);
if
( strlen(v25) <
=
48
)
v13
=
&aS_1[
10
*
v21
+
v9];
aS_1
S010010011
1100100100
0010111110
0110100100
0010010011
1101110101
0011110101
0110010101
0001001100
v13
=
&aS_1[
10
*
v21
+
v9];
aS_1
S010010011
1100100100
0010111110
0110100100
0010010011
1101110101
0011110101
0110010101
0001001100
v7
=
(v4
+
v5
/
6
)
%
6
;
v8
=
v5
+
v4;
v9
=
v22;
v20
=
v7;
v10
=
5
-
v8
%
6
;
for
( i
=
0
; ; i
=
1
)
{
switch ( v10 )
{
case
1
:
+
+
v9;
break
;
case
2
:
v17
=
(v21
+
+
&
1
)
=
=
0
;
v9
+
=
v17;
break
;
case
3
:
v12
=
(v21
+
+
&
1
) !
=
0
;
v9
-
=
v12;
break
;
case
4
:
-
-
v9;
break
;
case
5
:
v19
=
(v21
-
-
&
1
) !
=
0
;
v9
-
=
v19;
break
;
default:
v18
=
(v21
-
-
&
1
)
=
=
0
;
v9
+
=
v18;
break
;
}
if
( v9 >
9
)
break
;
if
( v21 >
8
)
break
;
v7
=
(v4
+
v5
/
6
)
%
6
;
v8
=
v5
+
v4;
v9
=
v22;
v20
=
v7;
v10
=
5
-
v8
%
6
;
for
( i
=
0
; ; i
=
1
)
{
switch ( v10 )
{
case
1
:
+
+
v9;
break
;
case
2
:
v17
=
(v21
+
+
&
1
)
=
=
0
;
v9
+
=
v17;
break
;
case
3
:
v12
=
(v21
+
+
&
1
) !
=
0
;
v9
-
=
v12;
break
;
case
4
:
-
-
v9;
break
;
case
5
:
v19
=
(v21
-
-
&
1
) !
=
0
;
v9
-
=
v19;
break
;
default:
v18
=
(v21
-
-
&
1
)
=
=
0
;
v9
+
=
v18;
break
;
}
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